Find all $c$ which satisfy $$\frac{c}{3} \le 2+c < -2(1+c).$$Express your answer in interval notation, simplifying any fractions which occur in your answer.
Solution: We have two inequalities which $c$ must satisfy. We consider these inequalities one at a time.

The first inequality is $\frac{c}{3}\le 2+c$. Multiplying both sides by $3$, we have $$c\le 6+3c.$$Subtracting $3c$ from both sides gives $$-2c\le 6.$$We can divide both sides by $-2$, but we must reverse the inequality since $-2$ is negative. This gives $c\ge -3$.

The second inequality is $2+c < -2(1+c)$. Expanding the right side, we have $$2+c < -2-2c.$$Adding $2c-2$ to both sides gives $$3c<-4.$$Dividing both sides by $3$ gives $c<-\frac{4}{3}$.

So, all $c$ which satisfy both inequalities are given by $-3\le c<-\frac{4}{3}$, or, in interval notation, $c\in\boxed{\left[-3,-\frac{4}{3}\right)}$.